[math] int_{0}^{frac{pi}{2}-1} sin(x+sin(x+...))dx [/math]

I got it. Before I type it out I just want to clarify that I am the best undergrad analyst in this board. Okay, let's get started.

Let [math] y = \sin(x + \sin(x + ...)) [/math]. Then clearly [math] y = \sin(x + y) [/math] and therefore [math] \arcsin(y) = x + y [/math] so [math] x = \arcsin(y) - y [/math].

We have accomplished two things. First, we isolated a variable. And second, we found a way to graph this function. See pic related to see the graph.

From the graph we can clearly see that:
[eqn] \int_{0}^{\frac{\pi}{2} - 1} y dx = \frac{\pi}{2} - 1 - \int_{0}^{1} xdy [/eqn]

Let's compute:
[eqn] \int_0^1 xdy = \int_0^1 \arcsin y - ydy = \frac{1}{2} (\pi - 3) [/eqn]

Therefore:
[eqn] \int_{0}^{\frac{\pi}{2}-1} ydx = \frac{\pi}{2} - 1 - \frac{1}{2}(\pi - 3) = \frac{1}{2} [/eqn]

This is the correct solution, good job user.

Can you guys figure out what the closed form of the middle intersection point is?

[math]

y:=\sin(x+\sin(x+\ldots)) \\
y=\sin(x+y) \\
y'= \cos(x+y) (1+y') \\
\cos(x+y) = \frac{y'}{1+y'} \\
\cos^2(x+y) = (\frac{y'}{1+y'})^2 \\
1-\cos^2(x+y)= 1-(\frac{y'}{1+y'})^2 \\
\sin^2(x+y) = \sqrt{1-(\frac{y'}{1+y'})^2} \\
\sin(x+y) = \sqrt{1-(\frac{y'}{1+y'})^2} \\
y=\sqrt{1-(\frac{y'}{1+y'})^2}

[/math]

Also,

[math]
\cos(x+y) = \cos x \cos y - \sin x \sin y = \cos x \cos y - y \sin x
[/math]

I don't fucking know man...

can you explain this step for a brainlet?

I am so fucking retarded...

Well, look at the graph. The part that is under the red curve is the area that we want to find. Well, that area is equal to the entire rectangle it is enclosed in minus the area that is above the curve (but inside of the rectangle). See pic related to see that the formula is

[math] A = \text{Rectangle} - B [/math].

And intuitively remember that if you take a graph in the xy plane and you rotate it 90 degrees you get a graph in the yx plane with the curve defined by the inverse of [math] y [/math]. And computationally to find the inverse of [math] y(x) [/math] you just isolate [math] x [/math] in terms of y, and we already had [math] x = \arcsiny - y [/math] so this was a viable route to take.

I'm going to admit that you never see problems like in the OP in a calc course, but when I did volume of revolution problems in Calc 2 we used and overused this technique.

Thanks. Don't remember ever having to deal with such a problem, but I'm just an engineering student.

Don't feel too bad. Before I got my solution I tried two things. First I tried doing what you did, get the derivative to see if I could then plug the resulting differential equation into wolfram alpha and get an explicit formula. (Always try this because if there is an explicit formula then why even bother with other methods). After wolfram told me no such expression existed I then graphed the function, noticed the symmetry at x=0 and then tried to see if I could do a substitution in the integral that would exploit this symmetry. After this failed me I finally came to the solution you see.

[math]

x=\arcsin(y) - y \\
dx = (\frac{1}{\sqrt{1-y^2}}-1) dy \\
x=0 \implies \arcsin(y) - y=0 \implies y=0 \\
x=\frac{\pi}{2}-1 \implies \arcsin(y) - y = \frac{\pi}{2}-1 \implies y=1 \\
\int_{0}^{\frac{\pi}{2}-1} y dx =\int_{0}^{1} y (\frac{1}{\sqrt{1-y^2}}-1) dy = 1- \frac{1}{2} = \frac{1}{2}

[/math]

I am just mad because I thought about using arcsin and saw that the domain allowed it, but I was trying other stuff at that time and then forgot about it..

>from the graph
brainlet

Question:
How do we know that this infinite composition of functions is integrable?
I found this:
en.wikipedia.org/wiki/Infinite_compositions_of_analytic_functions
but dunno....

In this case you can visually verify that it converges on the interval being integrated over.

fuck maths, literally only used basic counting and percentages and im successful as fuck

It is actually not as complicated as the link you posted it makes it seem. This infinite composition can be viewed as the limit of a sequence of functions [math] f_1 = sin(x), f_2 = sin(x + sin(x)), f_3 = sin(x + sin(x + sin(x))),... [/math] and the general rule is [math] f_n = sin(x + f_{n-1}) [/math]

And the theory of sequences of functions is pretty straight forward.

>And the theory of sequences of functions is pretty straight forward.
Is it though?

A sufficient condition would be that f_n converges uniformely in [0,π/2].
But how can this
[math] d(f_n,f_m):=\sup\limits_{x\in [0,\frac{\pi}{2}]}|f_n(x)-f_m(x)| \to 0 [/math]
be proven?
How do you write out [math] f_n(x)-f_m(x) [/math]?

I used |sin(x)|

it is easy to show that it is Lebesgue-integrable: One can use the Theorem of dominated convergence using the dominant function 1, and thats it. (since the sequence consists of integrable functions)

Oh I see. Thanks!
How does on prove pointwise convergence?

I am thinking Banach fixed point on using the metric space of all bounded real functions and the mapping T(f)=sin(id+f), but my brain is starting to melt.

No thoughts on this?

It looks to be the zero of cos(x)+2x, however I don’t think a closed form exists.

bump

[eqn]
u = x + \sin(x + \dots) \\
u = x + \sin u \\
du(1-\cos u) = dx
[/eqn]
the function [math]f(u) = u-\sin u [/math] is absol increasing in 0 to [math]\pi/2[/math] so the int limits after u sub are 0 and [math]\pi/2[/math]
[eqn]\int_0^{\frac{\pi}{2}} (1-\cos u)\sin u du = \left|-\cos u + \frac{\cos 2 u} {4}\right|_0^{\frac{\pi}{2}} = \frac{1}{2}[/eqn]

>Then clearly
F for not checking convergence

If the sequence of functions did not converge then OP would not ask me to find its integral, dumbass. If OP has asked to prove the function exist then I would have looked for ways to do that, but fortunately, I'm not retarded like you.

OP didn't ask for anything that makes sense actually, dumb nigger. Integrals can't be solved. So the point is moot and everything has to be checked accordingly.

God, I got up to [math]f(x)=\sin(x+f(x))[/math] but I was too much of a brainlet to think to isolate x instead of f(x).

>best analyst

you got cucked by on three lines

pseud

I didn't say that I was the best analyst. That's what the guy I replied to said.

this was meant for

It's not a competition on who gets the shortest solution lol

Yes, but you didn't even show pointwise convergence, mr. analyst. The banach theorem approach is correct though.

>mr. analyst
>tmw Veeky Forums posters can't take a little shade

We can't even get away with throwing some pointless banter even if we are the first solution now. I guess Veeky Forums is now a safespace for the brainlets.

getting away with banter on Veeky Forums is perfectly achievable, but it only works when one is smart you know

>from the graph we can see

holy shit, which shitshow do you attend that allows you steps like this haha

I got away with it, and people just kept the discussion normally. Commenting on my solution and posting new solutions. Until you came here and dropped the salt.

Anyways, I will say that I consider the two solutions posted after mine to be more elegant. But it is clear to anyone that all current solutions came out of the same foundation, and just take the idea in different directions.

If you are still a calculus student then yeah, you should not do that step. You could prove the identity I used rigorously by verifying a couple of inequalities, it is just that I don't really care about such a routine proof. This is similar to the people saying "you should prove the function converges". I'm solving a problem on an Indonesian cartoon connoisseur forum, not writing a paper. Why would I even bother with proving stuff that is obviously true instead of just solving the problem?

Anyways, given that these new posts made me come back here I got the idea of finding another tricky calculus problem and then making a thread about it. Be on the lookout for that.

BUMP

bad form 3/10

>not checking convergence
>"from the graph we can clearly see that"

one of the better threads in a while, gg op

BUMP

bump

Fuck off retard thread's over.

Thread is not over. Nobody proved that the integrand actually converges.

We need a proof that the sequence:
0,
sin(x + 0),
sin(x + sin(x)),
sin(x + sin(x + sin(x))),
...
is an increasing sequence that converges to a value y(x) such that y(x) = sin(x + y(x)), for 0 < x < (pi/2) - 1?

Proof by induction that this sequence is increasing and bounded above.
Let a(0,x) =0.
a(i,x) = sin(x + a(i-1,x)).
Clearly 0

You're welcome:
en.wikipedia.org/wiki/Integral_of_inverse_functions
In summary: if finding the integral of a certain function is too hard for you, check whether integrating the inverse function is easier, if so you can get the integral of the original function from the integral of its inverse

Proof: some geometric realizations about the areas between (a, f(a)) and (b, f(b)) and the axis (X and Y) at (a, 0), (b, 0), (0, f(a)) and (0, f(b)). Also consider the rectangle containing the points (0, 0), (a, 0), (0, f(a)). (a, f(a)).

thx, didn't know that theorem

>first discovered at 1905
>first proven without assumptions of differentiability at 1994
what the fuck

There's nothing to solve here, you brainlet.